The latent heat of vaporization of a liquid at $500 K$ and $1 a t m$ pressure is $10.0 k c a l / m o l$ . What will be the change in internal energy $(Δ U)$ of $3$ moles of liquid at the same temperature?

Question

Q. Latent heat of vaporization of a liquid at 500K and 1atm pressure is 10.0Kcal/mol. What will be the change in internal energy of 3 mol of liquid at same temperature and pressure?

Answer 1

3H2O(l)→3H2O(g)3H2O(l)→3H2O(g)

Δng=3Δng=3

ΔE=ΔH−ΔngRTΔE=ΔH−ΔngRT

=3×10−3×500×0.002=27Kcal=3×10−3×500×0.002=27Kcal

Here H is the enthalpy of vaporisation......

The latent heat of vaporization of a liquid at 500 K and 1 atm pressure is 10.0 kcal/mol. What will be the change in internal energy (ΔU) of 3 moles of liquid at the same temperature?

The correct option is C 27.0 kcal/mol3H2O(l)→3H2O(g)3H2O(l)→3H2O(g)Δng=3Δng=3ΔE=ΔH−ΔngRTΔE=ΔH−ΔngRT=3×10−3×500×0.002=27Kcal=3×10−3×500×0.002=27KcalHere H is the enthalpy of vaporisation......4.0

The latent heat of vaporization of a liquid at $500 K$ and $1 a t m$ pressure is $10.0 k c a l / m o l$ . What will be the change in internal energy $(Δ U)$ of $3$ moles of liquid at the same temperature?

The correct option is C $27.0 k c a l / m o l$
3H2O(l)→3H2O(g)3H2O(l)→3H2O(g)
Δng=3Δng=3
ΔE=ΔH−ΔngRTΔE=ΔH−ΔngRT
=3×10−3×500×0.002=27Kcal=3×10−3×500×0.002=27Kcal
Here H is the enthalpy of vaporisation......
4.0